Calorimetric Study of $HCl$ and $NaOH$
To determine the enthalpy change ($\Delta H$) of neutralization between a strong acid ($HCl$) and a strong base ($NaOH$).
Neutralization is an exothermic process. For strong electrolytes, the enthalpy is nearly constant ($\approx -57.3 \text{ kJ/mol}$) as it only involves the formation of $H_2O$.
$q = m \cdot c \cdot \Delta T$
$m = 100\text{g}, c = 4.18 \text{ J/g}\cdot\text{K}$
$m = V_{acid} + V_{base} = 50\text{mL} + 50\text{mL} = 100\text{mL}$
Density $\approx 1\text{g/mL} \implies$ $m = 100\text{g}$
$q = m \cdot c \cdot \Delta T$
$n = \text{Molarity} \times V(L)$
$n = 1.0\text{M} \times 0.050\text{L} = 0.050\text{ mol}$
$\Delta H = -q/n$
Calculated using $q/n$, where $n = 0.05\text{ moles}$ for $HCl + NaOH$.